If X is a Poisson variable such that $P(X=C)=2P(X=C+1).$ Then the variance of X is :

$\frac{C+1}{2}$

The correct answer is Option (2) → $\frac{C+1}{2}$

We are given that X follow a Poisson Distribution.

$P(X=k)=\frac{e^{-λ}λ^k}{k!}$

$P(X=C)=2P(X=C+1)$

$⇒\frac{e^{-λ}λ^C}{C!}=2×\frac{e^{-λ}λ^{C+1}}{(C+1)!}$

$⇒λ^C=2×\frac{λ^{C+1}}{C+1}$

$⇒λ=\frac{C+1}{2}$